Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`
Advertisements
उत्तर
We know `lim_(x -> 0) (sin x)/x` = 1
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = lim_(alpha -> 0) (sin (alpha^"n"))/(1/alpha^"n" (alpha^"n")) xx (alpha^(1/"m") * alpha^"m")/(sin alpha)^"m"`
= `lim_(alpha -> 0) alpha^"n" * (sin(alpha^"n"))/alpha^"n" xx 1/alpha^"m" * 1/(((sinalpha)^"m")/(alpha^"m"))`
= `lim_(alpha -> 0) alpha^("n" - "m") *(sin(alpha^"n"))/alpha^"n" xx 1/((sinalpha)/alpha)^"m"`
= `(lim_(alpha -> 0) alpha^("n" - "m")) xx (lim_(alpha^"n" -> 0) (sin(alpha^"n"))/alpha^"n") xx 1/(lim_(alpha -> 0) ((sin alpha)/alpha)^"m")`
Case (i) m = n
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("m" - "m")) (lim_(alpha^"m" -> 0) (sin(alpha^"m"))/alpha^"m") xx 1/(lim_(alpha -> 0) ((sin alpha)/alpha)^"m")`
= `(lim_(alpha -> 0) alpha^0) xx 1 xx 1/1`
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = 1 xx 1 xx 1` = 1
Case (ii) m > n then n – m < 0
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("n" - "m")) (lim_(apha^"m" -> 0) (sin(alpha^"m"))/alpha^"m") xx 1/((lim_(alpha-> 0) ((sin alpha)/alpha)^"m"))`
= `lim_(alpha -> 0) 1/(alpha^("m" - "n")) xx 1 xx 1/1`
= `oo xx 1 xx 1 = oo`
Since `lim_(alpha -> 0) 1/(alpha^("m" - "n")) = lim_(alpha -> 0) (1/0)^("m" - "n") = oo`
Case (iii) m < n then n – m > 0
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("n" - "m")) (lim_(alpha^"m" -> 0) (sin(alpha^"m"))/(alpha^"m")) xx 1/(lim_(alpha-> 0) ((sin alpha)/alpha)^"m"`
= `(0)^("n" - "m") xx 1 xx 1`
= 0
∴ `lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = {{:(1, "if", "m" = "n"),(oo, "if", "m" > "n"),(0, "if", m < n):}`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit :
`lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)`
| x | – 3.1 | – 3.01 | – 3.00 | – 2.999 | – 2.99 | – 2.9 |
| f(x) | – 0.24845 | – 0.24984 | – 0.24998 | – 0.25001 | – 0.25015 | – 0.25158 |
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 3) 1/(x - 3)`
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 0) sec x`
Evaluate the following limits:
`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + 1) - 1)/(sqrt(x^2 + 16) - 4)`
Evaluate the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1
Evaluate the following limits:
`lim_(x -> oo)(1 + "k"/x)^("m"/x)`
Evaluate the following limits:
`lim_(x -> 0) (3^x - 1)/(sqrt(x + 1) - 1)`
Choose the correct alternative:
`lim_(x -> oo) sinx/x`
Choose the correct alternative:
`lim_(x - pi/2) (2x - pi)/cos x`
Choose the correct alternative:
`lim_(x -> 0) sqrt(1 - cos 2x)/x`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
Choose the correct alternative:
`lim_(x -> 0) ("e"^(sin x) - 1)/x` =
Choose the correct alternative:
The value of `lim_(x -> 0) sinx/sqrt(x^2)` is
`lim_(x -> 0) ((2 + x)^5 - 2)/((2 + x)^3 - 2)` = ______.
